\(\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)

We have,
\(A=\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]\) 
\(\begin{array}{l} \Rightarrow A^{\prime}=\left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right] \end{array}\) 
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left\{\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]+\left[\begin{array}{ccc} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right]\right\}\) 
\(=\frac{1}{2}\left[\begin{array}{ccc} 4 & 11 & -5 \\ 11 & 6 & 3 \\ -5 & 3 & 8 \end{array}\right]=\left[\begin{array}{ccc} 2 & \frac{11}{2} & -\frac{5}{2} \\ \frac{11}{2} & 3 & \frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 4 \end{array}\right]\) 
which is symmetric matrix and
\(Q =\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left\{\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]-\left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right]\right\} \)
\(=\frac{1}{2}\left[\begin{array}{rrr} 0 & -3 & -7 \\ 3 & 0 & 7 \\ 7 & -7 & 0 \end{array}\right]=\left[\begin{array}{rrr} 0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0 \end{array}\right] \) 
which is skew-symmetric matrix
Now,\(P+Q=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)\) 
 \(=\left[\begin{array}{rrr} 2 & \frac{11}{2} & -\frac{5}{2} \\ \frac{11}{2} & 3 & \frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 4 \end{array}\right]+\left[\begin{array}{rrr} 0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0 \end{array}\right]=\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]=A\) 
Hence, A is represented as sum of symmetric and skew-symmetric matrix.

Hence, the given function is not differentiable at x=2.

(i) sin x, sin 2x are continuous, so f (x) is continuous in [0,\(\pi\)]
(ii) c= cos^-1 \(\left( \frac { 1\pm \sqrt { 33 } }{ 8 } \right) \)

\((I+A)^{3}=(I+A)(I+A)(I+A)\\ =\left(I^{2}+I A+A I+A^{2}\right)(I+A)=(I+A+A+A)(I+A)\\ =(I+3 A)(I+A) \quad\left[\because A^{2}=A\right]\)
\(=I^{2}+I A+3 A I+3 A^{2}\\=I+A+3 A+3 A\\=I+7 A\)

Consider 3(2X + 3 Y) - 2(3X + 2 Y)
\(=3\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-2\left[\begin{array}{rr} -2 & 2 \\ 1 & -5 \end{array}\right]\)
\(\Rightarrow 6 X+9 Y-6 X-4 Y=\left[\begin{array}{cc} 6 & 9 \\ 12 & 0 \end{array}\right]-\left[\begin{array}{rr} -4 & 4 \\ 2 & -10 \end{array}\right]\)
\(=\left[\begin{array}{cc} 6+4 & 9-4 \\ 12-2 & 0+10 \end{array}\right]=\left[\begin{array}{cc} 10 & 5 \\ 10 & 10 \end{array}\right]\)
\(\Rightarrow 5 Y=\left[\begin{array}{rr} 10 & 5 \\ 10 & 10 \end{array}\right] \)
\(\Rightarrow \ Y=\frac{1}{5}\left[\begin{array}{rr} 10 & 5 \\ 10 & 10 \end{array}\right]=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right]\)
\(\text { Now } \ 2 X =\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-3 Y=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-3\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right] \)
\(=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-\left[\begin{array}{ll} 6 & 3 \\ 6 & 6 \end{array}\right]=\left[\begin{array}{lr} 2-6 & 3-3 \\ 4-6 & 0-6 \end{array}\right] \)
\(=\left[\begin{array}{ll} -4 & 0 \\ -2 & -6 \end{array}\right] \)
\(\Rightarrow \ X=\frac{1}{2}\left[\begin{array}{rr} -4 & 0 \\ -2 & -6 \end{array}\right]=\left[\begin{array}{lr} -2 & 0 \\ -1 & -3 \end{array}\right]\)
\(\text {Hence, } \quad X=\left[\begin{array}{rr} -2 & 0 \\ -1 & -3 \end{array}\right] \text { ; }Y=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right]\)

\(\text { Let } E=\sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right.\)
\(\text { Now, } 2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right) \) 
\(\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right] \)
\(\Rightarrow \quad 2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{2 / 3}{8 / 9}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \)
\(\text { Put } \tan ^{-1}\left(\frac{3}{4}\right)=\theta \Rightarrow \tan \theta=\frac{3}{4} \) 
\(\text { Then, } \sin \theta=\frac{3}{5} \Rightarrow \theta=\sin ^{-1}\left(\frac{3}{5}\right) \)
\(\Rightarrow \tan ^{-1}\left(\frac{3}{4}\right)=\sin ^{-1}\left(\frac{3}{4}\right)\) 
From Eqs. (i) and (ii), we get
2\( \tan ^{-1}\left(\frac{1}{3}\right)=\sin ^{-1}\left(\frac{3}{5}\right)\)
Now, put \(\tan ^{-1} 2 \sqrt{2}=\phi \Rightarrow \tan \phi=2 \sqrt{2}\)
Then, \(\cos \phi=\frac{1}{3} \Rightarrow \phi=\cos ^{-1}\left(\frac{1}{3}\right)\)
Now, \(\tan ^{-1} 2 \sqrt{2}=\cos ^{-1}\left(\frac{1}{3}\right)\)
\(\tan ^{-1}\left(\frac{3}{4}\right)=\sin ^{-1}\left(\frac{3}{5}\right)\) 
\( \therefore E =\sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right) \)
\(=\sin \left(\sin ^{-1} \frac{3}{5}\right)+\cos \left(\cos ^{-1} \frac{1}{3}\right) \)
\(=\frac{3}{5}+\frac{1}{3}=\frac{9+5}{15}=\frac{14}{15} \)

We know tan x is increasing for \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)\(\left[y=\tan x \Rightarrow y^{\prime}=\sec ^{2} x>0\right]\)
\(\text {Also } 1>\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}>\tan \frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1} >1>\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}>1>\tan ^{-1}(1) . \)

\(\begin{array}{c|l|l|l|l} \hline x & 0 & 500 & 2000 & 3000 \\ \hline P(X) & \frac{9995}{10000} & \frac{3}{10000} & \frac{1}{10000} & \frac{1}{10000} \\ \hline P_{i} X_{1} & 0 & \frac{1500}{10000} & \frac{2000}{10000} & \frac{3000}{10000} \\ \hline \end{array}\) 
\(\because E(X)=\Sigma X P(X)\) 
= 0.65

Since, X = Number of four seen
On tossing two dice, X = 0, 1, 2
Also, \(P(4)=\frac{1}{10} \text { and } P(\operatorname{not} 4)=\frac{9}{10}\) 
So, \(P(X=0)=P(\operatorname{not} 4) \cdot P(\operatorname{not} 4)=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100}\) 
\(P(X=1)=P(\operatorname{not} 4) \cdot P(4)+P(4) \cdot P(\operatorname{not} 4)\) 
\(=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100}\) 
\(P(X=2)=P(4) \cdot P(4)=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100}\) 
Thus, we get the following table
\(\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & 81 / 100 & 18 / 100 & 1 / 100 \\ \hline \end{array}\) 
\(\begin{array}{c|c|c|c} \hline X P(X) & 0 & 18 / 100 & 2 / 100 \\ \hline X^{2} P(X) & 0 & 18 / 100 & 4 / 100 \\ \hline \end{array}\) 
\(\therefore \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}=\Sigma X^{2} P(X)-\left[\sum X P(X)\right]^{2}\) 
\(=\left(0+\frac{18}{100}+\frac{4}{100}\right)-\left(0+\frac{18}{100}+\frac{2}{100}\right)^{2}\) 
\(=\frac{22}{100}-\left(\frac{20}{100}\right)^{2}=\frac{11}{50}-\frac{1}{25}\) 
\(=\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18\)

\(\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\) 
\(\log \left|\frac{\sqrt{x-3}}{(x-1)^{1 / 6}(x+2)^{1 / 3}}\right|+C\)

(a) Let P be the given point (1, 6, 3) and M, the foot of perpendicular from P on the given line AB
  \(i.e.\quad \frac { x }{ 1 } =\frac { y-1 }{ 2 } =\frac { z-2 }{ 3 } (=k(say))\)

Any point on the given line is:
(k , 1+ 2k , 2 + 3k).
For some value of k, let the point be M.
:. Direction-ratios of PM are:
< k - 1 , 1 + 2k - 6, 2 + 3k - 3 >
i.e. < k - 1, 2k - 5, 3k - 1 > .
Since \(PM\bot AB\)  ,
:. (1) (k-l) + (2) (2k- 5) + (3) (3k-l) = 0
\(\Rightarrow\)k - 1 + 4k - 10 + 9k - 3 = 0
\(\Rightarrow\) 14k = 14 \(\Rightarrow\) k = 1.
:. Foot of perpendicular Mis (1,1+2,2+3) i.e. (1,3,5).
Let \(P'(\alpha ,\beta ,\gamma )\)  be the image of P in the given line.
Then M is the mid-point of [PP']
\(\frac { \alpha +1 }{ 2 } =1\frac { \beta +6 }{ 2 } =3,\frac { \gamma +3 }{ 2 } =5\)
\(\alpha +1=2,\beta +6=6,\gamma +3=10\)
\(\Rightarrow \alpha =1,\beta =0,\gamma =7\)
Hence the reqd image is(1,0,7).
(b) (i) The equations of PP' are:
\(\frac { x-1 }{ 1-1 } =\frac { y-6 }{ 0-6 } =\frac { z-3 }{ 7-3 } \)
\(i.e.\quad \frac { x-1 }{ 0 } =\frac { y-6 }{ -6 } =\frac { z-3 }{ 4 } \)
\(\Rightarrow \frac { x-1 }{ 1-1 } =\frac { y-6 }{ -3 } =\frac { z-3 }{ 2 } \)
(ii) Length od segment [PP']
\(\sqrt { { \left( 1-1 \right) }^{ 2 }+{ \left( 0-6 \right) }^{ 2 }+{ \left( 7-3 \right) }^{ 2 } } \)
\(=\sqrt { 0+36+16 } =\sqrt { 52 } =2\sqrt { 13 } units.\)

Since \(l, m, n \text { and } l+\delta l, m+\delta m, n+\delta n\) are direction cosines of a variable line in two different positions,therefore,
\(l^{2}+m^{2}+n^{2}=1\)
\(\text { and }(l+\delta l)^{2}+(m+\delta m)^{2}+(n+\delta n)^{2}=1\)
\(\text { Now, }(l+\delta l)^{2}+(m+\delta m)^{2}+(n+\delta n)^{2}=1\)
\(\Rightarrow\left(l^{2}+m^{2}+n^{2}\right)+2(l \cdot \delta l+m \cdot \delta m+n \cdot \delta n) \)
\(+(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}=1[\text { from }(i i)] \)
\(\Rightarrow 1+2(l \delta l+m \delta m+n \delta n)+(\delta l)^{2} \)
\(+(\delta m)^{2}+(\delta n)^{2}=1[\text { using }(i)] \)
\(\Rightarrow 2(l \delta l+m \delta m+n \delta n)=-\left[(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\right] \)
\(\Rightarrow(l \delta l+m \delta m+n \delta n)=-\frac{1}{2}\left[(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\right] \)
\(\text { Also } \cos (\delta \theta)=l(l+\delta l)+m(m+\delta m)+n(n+\delta) \)
\(=\left(l^{2}+m^{2}+n^{2}\right)+(l \delta l+m \delta m+n \delta n) \)
\(=1-\frac{1}{2}\left[(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\right] \quad[\text { using }(i) \text { and (iii)] }\)
\(\Rightarrow 2(1-\cos \delta \theta)=(\delta I)^{2}+(\delta m)^{2}+(\delta n)^{2}\)
\(\Rightarrow 2 \times 2 \sin ^{2} \frac{\delta \theta}{2}=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\)
\(\left[\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right]\)
\(\Rightarrow 4\left[\frac{\delta \theta}{2}\right]^{2}=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}\)
\(\text { [if } \theta \text { is small, then } \sin \theta \rightarrow \theta]\)
\(\Rightarrow(\delta \theta)^{2}=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2} \)
which is true.

\(\Rightarrow \)  y - x + log |x + y| = c is the required solution.

The function y = (sin x)^{sin x} is defined for all positive real numbers. Taking logarithms, we have
log y = log (sin x)^{sin x} = sin x log (sin x)
Then \(\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}(\sin x \log (\sin x))\)
\( =\cos x \log (\sin x)+\sin x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x) \)
\(=\cos x \log (\sin x)+\cos x \)
\(=(1+\log (\sin x)) \cos x\)
Thus, \(\frac{d y}{d x}=y((1+\log (\sin x)) \cos x)=(1+\log (\sin x))(\sin x)^{\sin x} \cos x\)

Find point of intersection on two curves and show that slopes of tangents to the two curves at point of intersection is same.

\({2\over3}sin^{-1}\sqrt{x^3\over a^3}+c\)

\({x\over a}+{y\over b}=1\)

\(2-\left[ \frac { 1+\sqrt { { 1+x }^{ 2 } } }{ 1+{ x }^{ 2 } } \right] >0\ as \ \frac { 1+\sqrt { { 1+x }^{ 2 } } }{ 1+{ x }^{ 2 } } <\quad 2\).

\(c^3\over6\sqrt{3}\) cu units